$u=2xy, v=x^2-y^2 \cdots (Given)$ $\dfrac {\delta (u,v)}{\delta (r,\theta)}= \dfrac {\delta (u,v)}{\delta (x,y)}. \dfrac {\delta (x,y)}{\delta (r,\theta)} \cdots(A)$ $\dfrac {\delta (u,v)}{\delta (x,y)}= \begin{bmatrix} \dfrac {\delta u}{\delta x} &\dfrac {\delta u}{\delta y} \[0.3em] \dfrac {\delta v}{\delta x} & \dfrac {\delta v}{\delta y} \[0.3em]

\end{bmatrix} $ $\dfrac {\delta u}{\delta x}=2y;\ \dfrac {\delta u}{\delta y}=2x;\ \dfrac {\delta v}{\delta x}=2x; \dfrac {\delta v}{\delta y}=-2y$ $\dfrac {\delta (u,v)}{\delta (x,y)}= \begin{bmatrix} 2y&2x \[0.3em] 2x & -2y \[0.3em] \end{bmatrix} =\ -4y^2-4x^2,\ = - 4(x^2+y^2) \cdots (1) $ $\dfrac {\delta (x,y)}{\delta (r,\theta)}= \begin{bmatrix} \dfrac {\delta x}{\delta r} &\dfrac {\delta x}{\delta \theta} \[0.3em] \dfrac {\delta y}{\delta r} & \dfrac {\delta y}{\delta \theta} \[0.3em]

\end{bmatrix} $ $\dfrac {\delta x}{\delta r}=\cos\theta;\ \dfrac {\delta x}{\delta \theta}=-\sin \theta (r);\ =\ -r\sin \theta;\ \dfrac {\delta y}{\delta r}=\sin \theta;\ \dfrac {\delta y}{\delta r}=\cos \theta (r)=\ r \cos \theta$ $\dfrac {\delta (x,y)}{\delta (r,\theta)}= \begin{bmatrix} \cos \theta& -r \sin \theta \[0.3em] \sin \theta& r \cos \theta \[0.3em]

\end{bmatrix} = r \cos^2 \theta + r \sin^2 \theta =\ r \cdots (2) $ Substitute equation 1, 2 in (A), we get, $\dfrac {\delta (u,v)}{\delta (r,\theta)}=-4 [r^2 \cos^2 \theta + r^2 \sin^2 \theta] \times r$ $-4 [r^2 (\cos^2 \theta + \sin^2 \theta)] \times r=\ -4r^3$ $\dfrac {\delta (u,v)}{\delta (r,\theta)}=-4r^3$